Question: Let $g(x)=\sqrt[3]{x^4}$. $g'(x)=$
Solution: The strategy We can first rewrite $g(x)$ as a rational power of $x$. Then, the derivative of $g$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Rewriting the radical as a rational power $\begin{aligned} g(x)&=\sqrt[3]{x^4} \\\\ &=x^{^{\frac{4}{3}}} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{4}{3}}}\right) \\\\ &=\dfrac{4}{3}x^{^{\frac{4}{3}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac43x^{^{\frac{1}{3}}} \end{aligned}$ In conclusion, we found that $g'(x)=\dfrac43x^{^{\frac{1}{3}}}$. This can also be written as $\dfrac43\sqrt[3]{x}$ (all equivalent forms are accepted).